Solving a simple logarithmic equation with a log on both sides. For this example what we have is a single log is equal to a single log. And these are actually really quite easy to solve because if you think about it, if we have the log base to the same thing, what’s in the inside has to be the same as well. There’s no way to make these two logs equal to each other unless the insides are the same also.

So what we can actually do is just drop our logarithms all together and we know that 5 minus x has to be equal 8, the insides have to be the same. Solving this out, solve it, you could either bring the x over or the 5 over divide, doesn’t really matter, just solving this out. I’m going to bring the 5 over so we get -x is equal to, subtract 5 is equal to 3 and x is equal to -3.

We did say earlier that our answers can be negative but what we have to be careful of is making sure that when we plug it back in, we have a positive number. So we take our x is -3, we plug this back in we end up with a log of a positive number so it actually works. We can get negative answers we just have to make sure that when we plug them in we end up with a positive number inside of our log. So this actually works out to be our answer.

How this actually works is there was a property if you remember that was said b to the log base b of x is equal to x, if our exponential base and our logarithm base is the same and this is x, what we are actually doing in this case, if you actually want to see the work behind it, is we’re taking 7 to both sides here. So a little bit more back ground is, you could just say the insides have to be the same but what we’re really doing is taking 7 to the power of both of these, then we have this formula for both cases, 7 to the log base 7, those cancel leaving us with x minus 5. 7 to the log 7 again those cancel leaving us with 8.

So the easy way of looking at it is just saying okay the insides have to be the same, but what we’re really doing is using that property of log to get rid of our logarithms.